# Tag Info

### What is a Haar random quantum state?

Typically this is a slight abuse of notation. One can have a unitary operator $U$ chosen from some Haar measure, such as the circular unitary ensemble. Then, taking some fiducial state $|\psi_0\rangle$...
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### Is there a lower bound on the average diamond norm of two uniformly random unitaries U1 and U1 of dimension D that are sampled from haar measure?

This answer won't actually give you a bound, but will provide some information that may help you in your search. You may be able to find an answer in the random matrix theory literature if you ...

### Expected value of a Haar random quantum state multiplied by a unitary

I'm writing an alternate proof because it uses some interesting tools, computes the value of these expressions, and gives some insights into how we can interpret the quantities in consideration. The ...
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### Multiplication by a Haar random unitary two times

There is an explicit formula for the integral with respect to the Haar measure of any polynomial in the entries of a unitary and its conjugate, due to Collins and Śniady: Benoît Collins and Piotr ...
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### Confusion about the output distribution of Haar random quantum states

The two facts are connected in that they both arise as a result of rotational invariance of the Haar measure. We will derive them in the case of large $n$ since this is when the Porter-Thomas ...
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### Random quantum states and Schur-Weyl duality

Note that the quoted relation $$\bar M_i = \sum_\lambda a_\lambda P_\lambda,$$ only holds if the $M_i$ also commute with the representation of the symmetric group! Otherwise this can obviously not ...

### What is the expectation value ${\Bbb E}[\langle\psi,O\psi\rangle]$ over the Haar distribution?

Since a Haar-random $\lvert\psi\rangle=U\lvert0\rangle$ for a Haar-random $U$, your expectation value equals $$\langle 0 \rvert \Big[\int \mathrm d U\, UOU^\dagger\Big]\lvert0\rangle\ .$$ The ...

### Is the Haar measure invariant under conjugation?

I will answer this question in a more general context. You might know that Haar's theorem tells you that on any locally compact group $G$, there is a unique left-invariant (Borel) measure $\mu$, up to ...
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### Compute the large $n$ distribution of $|\langle z_i|\psi\rangle|^2$ over Haar random quantum states

In the following, I'll show the evaluation of the probability densities of the transition probabilities: $|\langle \psi | z\rangle^2$ and their pairwise independence. I didn't work out the full mutual ...
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### Computing expectation value of $|\langle z|C|0^n\rangle|^2$ over Haar random circuit

The issue that easily leads to confusion is the dual role played by output bitstring probability. It enters the computation of the average in two ways. On one hand, it determines how often one sees ...

### Is there a lower bound on the average diamond norm of two uniformly random unitaries U1 and U1 of dimension D that are sampled from haar measure?

The result you're looking for is effectively Proposition 19 of the paper: Almost all quantum channels are equidistant; which I'm rewriting here for convenience: Let $U, V \in \mathcal{U}(d)$ be two ...
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### Generating random, but non-uniform state

Rejection sampling is a good fit and works without any changes, simply by plugging the desired distribution $p(\psi)$ into the standard algorithm. Let$^1$ $M:=\max_{\psi\in\mathbb{CP}^1} p(\psi)$. To ...
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### What is the expectation value of $|\langle \psi|U|\psi \rangle|$ over Haar random states $|\psi\rangle$?

This partial answer calculates the integral for $d=2$. In this case, every traceless unitary $U$ is equivalent to the Pauli $Z$ up to similarity and global phase, so, by rotational invariance of the ...
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### Is the column vector of a uniformly sampled random unitary matrix a uniformly sampled random state vector?

Suppose that was not the case. Then taking the first column of a uniformly random unitary matrix gives you a nonuniformly random state. That means that there is some state, call it $|\psi\rangle$, ...

### Is the column vector of a uniformly sampled random unitary matrix a uniformly sampled random state vector?

Yes. A uniformly (Haar random) sampled state vector $|\psi\rangle$ is characterized by the fact that the probability measure is invariant under any $U$, i.e., colloquially, $U|\psi\rangle$ is just as ...

All you need are simple tools from measure concentration. The setup is as follows (repeated from the question above for completeness): $| \psi \rangle$ is an $n$-qubit state and $| \alpha \rangle := | ... 3 votes ### On the distribution of the fidelity of a random product state with an arbitrary many-qubit state The required fidelity$F$is a function of the Cartesian product of the single$n$-qubit state space:$CP^{2^n-1}$and$n$copies of a single qubit state space:$CP^{1} \cong S^2$. The statistics ... 3 votes ### What is the probability$\Pr(\|U-I\|_{\rm op}<\varepsilon)$of a Haar-random unitary being close to the identity?$U-I$is a normal matrix so$||U-I||_{op}$is its eigenvalue with the largest magnitude. The eigenvalue equation for this matrix is $$(U-I)|\psi\rangle=\lambda|\psi\rangle,$$ so $$|\lambda|^2=(\cos\... 3 votes Accepted ### Quantum hardness of XQUATH conjecture Maybe think of it this way - a quantum computer, executing a small enough random circuit C acting on a state initially prepared as \vert 0^n\rangle and sampling therefrom, will get an n-bit ... 3 votes Accepted ### Quantum supremacy: shallow depth Haar random circuits and unitary designs First of all, that does not imply anything for shorter (constant/logarithmic) depths. Moreover, the 2-design property does not imply that the outcome distribution is the same as for Haar-random ... 3 votes ### Spoofing XQUATH with the Feynman method The paper does not specify the exact algorithm or class of distributions \mathcal{D} for which such algorithm fails to refute XQUATH, and some classes of distributions \mathcal{D} do not satisfy ... 3 votes ### Prove that uniformly random states have moments {\bf E}_\psi|\langle x|\psi\rangle|^{2t}\sim1/\binom d t The factor in your claim is wrong. It should be \binom{d+t-1}{t}^{-1}. The correct claim follows from the identity$$ \int |\psi\rangle\langle\psi|^{\otimes t} d\psi = \binom{d+t-1}{t}^{-1} P_{\... 3 votes Accepted ### Density matrices of multiples copies of a single Haar-Random state You can think of this as a generalization of the maybe better-known result that $$\int d\psi \,\mathbb{P}_\psi = \frac{I}{d},$$ where I used the notation$\mathbb{P}_\psi\equiv|\psi\rangle\!\langle\...
Intuitively, that's the case: the vector being random, there is no reason to prefer $|0\rangle$ over $|1\rangle$ on the first qubit. I don't think it requires to compute some integrals other than ...